Sketch $x^2 e^{-x}$

Question

In my interview to join a university to study physics, I was asked to sketch $y = x^2 e^{-x}$ at the time I could not do so. The interviewer told me that I need to have mathematical intuition like this to study physics. Now, I am studying physics at the university buy I still can not visualize the equation.

I could just sketch it by differentiating the entire equation or by substituting in values but I do not want to do this since I do not think this is what the interviewer wanted me to do; it does not really give intuition for the shape.

https://www.desmos.com/calculator/u0yt84tpub[what the equation looks ljke][1] [What the two parts of the product looks like individually][2] The negative part of $y$ is very easy to see: both $x^2$ and $e^{-x}$ are positive, and they both increase when $x$ become more negative so $y$ becomes larger as $x$ becomes more negative.

The positive part I still can not explain. I've tried many approach but I could not find an intuitive way to explain the shape. I didn't really have anyone around me who I can ask so I asked here.

Thank you very much for a quick and kind response. I appreciate them very much In addition, to the main question I was wondering how I could build a stronger intuitions in mathematics. I understand this is a difficult question to answer. Infact my interviewer said that the reason why he tests peoples intuition is because he believes that it is something that it cant be taught. I think this kind of intuition is a weak point in my study. I want to know some advice on what kind of training I could do to build stronger intuition. — Wanting to be anAndroidDevelor, Sep 01 '18 at 14:07

score 3 · Accepted Answer · answered Sep 01 '18 at 13:05

3

When $x$ is positive but small, the dominant part is $x^2$ and the shape is roughly that.

However as $x$ is very big, $\exp(-x)$ will have a bigger influence and it will pull down the whole curve.

Polynomial has bigger impact when $x$ is small and exponential will have a bigger impact will $x$ is large.

answered Sep 01 '18 at 13:05

Siong Thye Goh

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What would I say to back up the fact "Polynomial has bigger impact when x is small and exponential will have a bigger impact will x is large" with evidence. I think this is quite easy to do if the relation is y = a + b but I could not find a simple way to do this when the relation is y = a * b – Wanting to be anAndroidDevelor Sep 01 '18 at 14:09
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$\lim_{x \to 0} \frac{\exp(-x)}{x^r}=\infty$ – Siong Thye Goh Sep 01 '18 at 14:21
Oh wow this is very true thank you very much – Wanting to be anAndroidDevelor Sep 01 '18 at 14:25
I'm sorry for asking to be spoon fead but how would I show that x^2 is dominant over e^-x when x is small without substitution or differentiation? – Wanting to be anAndroidDevelor Sep 01 '18 at 17:01
Actually I think I can just look at the same rquation exp(0)/(x^r) for small value of x and shows that x^r is dominant – Wanting to be anAndroidDevelor Sep 01 '18 at 17:08
Actually the above approach doesn't really work since the equation is infinite around x = 0. But I can say the derivative of the equation is negative around 0 therefore the "dominantness" of the exp(-x) is decreasing so you get a hump like shape but again maybe this would not be as intuitive – Wanting to be anAndroidDevelor Sep 01 '18 at 17:31

score 1 · Answer 2 · answered Sep 01 '18 at 13:07

Let's deal with the non-negative part.

Differentiating is what you have to do. We have $$\frac{dy}{dx}=2xe^{-x}-x^2e^{-x}=xe^{-x}(2-x)$$ and to find critical points, set it to zero to give $x=0,2$ as $\exp$ is never zero.

Now at $x=0$, $y=0$ and at $x=2$, $y=4/e^2$. However, as $\exp$ grows faster than $x^2$, for large $x$, $y$ will tend to zero.

Putting this altogether, we get this graph.

score 0 · Answer 3 · answered Sep 01 '18 at 13:15

As you say, the part where $x \leq 0$ is simple.

For $x >0$, consider $$f(x)=x^2 e^{-x} \implies f'(x)=-e^{-x} (x-2) x\implies f''(x)=e^{-x} (x^2-4x+2)$$ So, the first derivative cancels at $x=0$ and $x=2$ and the second derivative test shows that the first is a minimum while the second is a maximum.

Moreover, when $x$ becomes large, the value of the function is smaller and smaller. I am sure that use that, you are abale to "see" what is the sketch of the function.

Sketch $x^2 e^{-x}$

3 Answers3